If out of 20 consecutive whole numbers two are chosen at random, then find the probability that their sum is odd.

Explanation

To solve the problem of finding the probability that the sum of two randomly chosen numbers from 20 consecutive whole numbers is odd, we can follow these steps:

### Step 1: Identify the Total Number of Ways to Choose Two Numbers
We need to calculate the total number of ways to choose 2 numbers from 20 consecutive whole numbers. This can be done using the combination formula:

\[ \text{Total ways} = \binom{20}{2} = \frac{20!}{2!(20-2)!} = \frac{20!}{2! \cdot 18!} \]

### Step 2: Simplify the Combination
Now we simplify the combination:

\[ \binom{20}{2} = \frac{20 \times 19 \times 18!}{2 \times 1 \times 18!} = \frac{20 \times 19}{2 \times 1} = 10 \times 19 = 190 \]

### Step 3: Identify Favorable Outcomes for Odd Sum
To have an odd sum, we need to choose one even number and one odd number. In 20 consecutive whole numbers, there are exactly 10 even numbers and 10 odd numbers.

### Step 4: Calculate the Number of Favorable Outcomes
The number of ways to choose 1 even number from 10 even numbers and 1 odd number from 10 odd numbers is:

\[ \text{Favorable outcomes} = \binom{10}{1} \times \binom{10}{1} = 10 \times 10 = 100 \]

### Step 5: Calculate the Probability
Now we can find the probability that the sum of the two chosen numbers is odd:

\[ \text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{100}{190} = \frac{10}{19} \]

### Final Answer
Thus, the probability that the sum of two randomly chosen numbers from 20 consecutive whole numbers is odd is:

\[ \frac{10}{19} \]

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